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Rings Over Which Every Simple Module Is Rationally Complete

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Rings Over Which Every Simple Module Is Rationally Complete Can. J. Math., Vol. XXV, No. 4, 1973, pp. 693-701 RINGS OVER WHICH EVERY SIMPLE MODULE IS RATIONALLY COMPLETE S. H. BROWN 1. Introduction. In 1958, G. D. Findlay and J. Lambek denned a relation­ ship between three i^-modules, A ^ B(C), to mean that A C B and every i^-homomorphism from A into C can be uniquely extended to an irreducible partial homomorphism from B into C. If A ^ B(B), then B is called a rational extension of A and in [5] it is shown that every module has a maximal rational extension in its injective hull which is unique up to isomorphism. A module is called rationally complete provided it has no proper rational extension. In 1969, R. C. Courter [1] completely characterized the rings for which every module is rationally complete. Furthermore, the rings for which every right i^-module is rationally complete are described in the review of [1] which appeared in Zbl. 182, p. 55. This paper is concerned with the related problem of describing the rings for which every simple right module is rationally complete. It is shown that the collection of rings which has this property contains all finite direct sums of matrix rings over duo rings. It is also shown that the ring [? a does not belong to this collection. Results are also obtained concerning rings for which every simple right module is injective. Acknowledgement. The author wishes to thank the referee for helpful suggestions. 2. Preliminaries. Throughout this paper, R will denote an associative ring with unity 1 and each module M will be a unitary right i^-module. E (M) will denote the injective envelope of M. 2.1 Definition. If A and B are two i£-modules and C C B, then / 6 HomB(C, A) is called a partial homomorphism of B into A. The set of all feHomB(C,A) for any C^B will be denoted by ParB(B,A). If / G ParR(B, A) and/ cannot be extended to another partial homomorphism of B into A whose domain properly contains the domain of /, then / will be called an irreducible partial homomorphism of B into A. 2.2 Definition. In [5] Findlay and Lambek have defined the following relationship between A, B and C: If A Ç B, then A ^ B{C) means that for Received February 21, 1972 and in revised form, May 24, 1972. 693 Downloaded from https://www.cambridge.org/core. 25 Sep 2021 at 18:58:07, subject to the Cambridge Core terms of use. 694 S. H. BROWN any i^-homomorphism / G Par^CB, C), f(A) = 0 if and only if / is a zero mapping. In general, if A = D and D g B(C), we identify A with D and writer g>B(C). The following four propositions are taken from [5] and the reader is referred to that paper for their proofs. 2.3 PROPOSITION. If C ^ B(A) and C CD QB then C g D{A) and D g 5(^). 2.4 PROPOSITION. // C :§ -5(4) anrf T/' is a?ry homomorphism into B, then +-HC) Sr1(B)(A). 2.5 Definition. B is said to be a rational extension of A provided that A ^ B(B). A module is called rationally complete if and only if it has no proper rational extension. 2.6 Definition. A submodule N of M is called large in M (written N Q' M) and M is called an essential extension of N provided that N P\ K ^ 0 for every nonzero submodule K of M. 2.7 PROPOSITION. (i) A ^ B(B) implies A Q' B. (ii) If A Q'A*, then C S B(A) if and only if C S B(A*). 2.8 PROPOSITION. A g B{B) if and only if for all bh 0 j£ b2 G B, there exists r G R such that b±r G A and bit ^ 0. 2.9 PROPOSITION. Let M be any module, let E = E(M), let A = HomB(£, E) and let ML = {X G A|X(M) = 0}. Then C(M) = fl {kerX|X G AfA} w a maximal rational extension of M which is contained in E. If G is any other maximal rational extension of M, then the identity mapping of M into C(M) can be extended to an isomorphism of G onto C(M). Proof. See [3, p. 60]. 2.10 Notation. Let A and B be two i^-modules so that A C B. If b G B, E (A : b)R = {r G R\br G A} and (b) = {r G i?|6r = 0}. 2.11 PROPOSITION. Let A be any module, E = E(A) and C = C(A) (as in 2.9). Then C = {x G £|(i4 :x)fi £ R(A)\. Proof. Let x G E be such that (4 : x)R ^ R(A) and let X G HomB(£, E) B be such that \(A) = 0. Then (4 : x)B C (X(x)) so that by 2.3, B fî (X(x)) g R(A) and hence (X(x)) g i?(£) by 2.7(h). Define/ G HomB(R,E) by /(f) = X(x)r for all r £ R. Then /((X (*))*) = 0. Therefore image / = 0 so that f{\) = X(s)l = 0. Thus X(x) = 0. Hence x G ker X for all X G UomB(E, E) such that X(4) = 0 and thus x G C. Downloaded from https://www.cambridge.org/core. 25 Sep 2021 at 18:58:07, subject to the Cambridge Core terms of use. SIMPLE MODULE 695 Ux Ç C, then 4 S C(C) so that A S (A + xR) (A). Define if/:R~>A + xR 1 by if/(r) = xr for ail r G R. Then i/^U) = (4 : x)B and i^- ^ + xR) = i?. From 2.4, it follows that (4 : x)R ^ i?G4). R 2.12 Definition. Let M be any module. Then ZR(M) = {m e M\ (m) C' £} is called the singular submodule of M. 2.13 PROPOSITION. If ZR(M) = 0, then M ^ N(N) if and only if M Q' N. Proof. See [3, Chapter 7]. 2.14 Remark. M Ç C(M) C E(jfcf) and it follows from 2.13 that if Z«(Af) = 0, then C(M) = E(M). 3. «-Rings. 3.1 Definition. A module will be called a simple module provided it contains no proper submodules. 3.2 Remark. As an immediate consequence of the definition, if A is a simple module, then: (i) If 0 5* x G £04) then A C' x£. (ii) ^4 ~ R/M where If is a maximal right ideal in R. The following question is currently an unsolved problem: What are the rings for which every simple module is injective? In [8], Rosenberg and Zelinsky credit Kaplansky with the partial solution below: 3.3 THEOREM. If R is a commutative ring with 1, then every simple R-module A is injective if and only if for every x Ç R there is a y £ R so that x2y = x. In this paper this result is extended to a larger class of rings which includes the commutative rings with 1. Since every injective module is rationally complete, it seems natural to consider the related question: What are the rings for which every simple module is rationally complete? 3.4 Definition. If R is a ring such that every simple module is rationally complete, then R will be called an a-ring. The symbol^ will be used to denote the collection of all a-rings. 3.5 Example. Let D be any division ring and let R = n n and M = 0 DI . Then for 0 ^ m G M", m = Mj ^ for some d G D, d 9* 0. Thus mR = [l o][o D] = M Downloaded from https://www.cambridge.org/core. 25 Sep 2021 at 18:58:07, subject to the Cambridge Core terms of use. 696 S. H. BROWN so that Af is a simple module. If 0 7^ ra = \ £ M, then 0 = [0 oC] and hence {m)B = [Do 0 ] • Now R 0 ] n [0 D] = °so that R (m ) is not large in R and hence ZB(M) = 0. Let # = Rj ^1 • Then if Vdi d2l [0 u\ _ [~0 diw + d2vl Lo 0 J ' TO wj " Lo 0 J e so if x ^ 0, then either d\ or d2 9 0 so w and z; can be chosen so that diu + d2v 9* 0 and hence M H xR 9* 0. Thus M Q' N. Then by 2.14, since ZB(M) = 0, M <* N(N). Therefore M is a simple i^-module which is not rationally complete. 3.6 Remark. Example 3.5 shows that there exists at least one non-trivial example of a ring which is not an a-ring. Since every division ring is a member oîs/,s/ is not an empty collection. 3.7 PROPOSITION. If R £S/ and if f is any ring homomorphism on R, then f(R) e sf. Proof. Let/ be a ring homomorphism and let T = f(R). If AT is any right T module, then by defining a • r = af(r) for all r G R, a G A, A becomes a right i^-module and if A T is simple, so is AB. Now if B is any right T-module so that A g B(BT), then by 2.8, if 61, 62 G 5 and 62 ^ 0, there is a / G T so e that bit G ^4 and b2t 9 - 0.
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